Angular width of the central maxima in the Fraunhofer diffraction for $\lambda$ = 6000 � is $\theta_{0}$. When the same slit is illuminated by another monochromatic light, the angular width decreases by 30%. The wavelength of this light is,

Question

In a single slit diffraction pattern, a light of wavelength $ 6000 \, \text{Å} $ is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from the slits. The width of the slit is ______ $ \times 10^{-4} $ m.

Answer 1

To find the wavelength of the light when the angular width of the central maxima decreases by 30%, we can use the formula for the angular width of the central maxima in Fraunhofer diffraction.

The angular width \theta_0 of the central maxima is given by:

\theta_0 = \frac{2\lambda}{a}

where \lambda is the wavelength of the light used and a is the width of the slit.

Given:

Initial wavelength \lambda_1 = 6000 \, \text{\AA}
With this wavelength, the angular width is \theta_0.
New angular width after decrease, \theta = 0.7 \theta_0 (since it decreases by 30%).

Let's say the new wavelength is \lambda_2. The formula for the new angular width is:

\theta = \frac{2\lambda_2}{a}

Since the angular width decreases by 30%:

\frac{2\lambda_2}{a} = 0.7 \left(\frac{2\lambda_1}{a}\right)

Cancel out common terms:

\lambda_2 = 0.7 \lambda_1

Substitute \lambda_1 = 6000 \, \text{\AA}:

\lambda_2 = 0.7 \times 6000 \, \text{\AA} = 4200 \, \text{\AA}

Thus, the wavelength of this light is 4200 \, \text{\AA}.

Therefore, the correct answer is 4200 �.

Answer 2